Number of positive integers x satisfying equation frac{1}{x}+frac{1}{x+1}+frac{1}{x+2}=frac{13}{2} i

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4-2.Quadratic Equations and Inequations

hard

The number of positive integers $x$ satisfying the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}=\frac{13}{2}$ is.

A

$0$

B

$1$

C

$2$

D

more than $2$

(KVPY-2021)

Solution

(c)

$\frac{1}{ x }+\frac{1}{ x +1}+\frac{1}{ x +2}=\frac{13}{12}$

$\frac{ x ^2+3 x +2+ x ^2+2 x + x ^2+ x }{ x ( x +1)( x +2)}=\frac{13}{12}$

$\frac{3 x ^2+6 x +2}{ x ( x +1)( x +2)}=\frac{13}{12}$

$36 x ^2+72 x +24=13\left( x ^3+3 x ^2+2 x \right)$

$13 x ^3+3 x ^2-36 x -24=0$

$( x -2)\left(13 x ^2+29 x +12\right)=0$

only one positive integral value $x =2$

Standard 11

Mathematics

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